• Lecture 3: Resistance is futile
ECEN 1400 Introduction to Analog and Digital Electronics
Lecture 3
Resistors
•
•
•
•
•
A note on charge carriers
Physics – origin of resistance
Ohm’s Law
Power dissipation in a resistor
Combinations of resistors
–  Series = voltage divider
•  The potentiometer
–  Parallel = current divider
•  Input and output resistance
•  Clicker quizzes
An alternative to the hydraulics analogy…
Robert R. McLeod, University of Colorado
25
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Charge carriers
•  Due to an unfortunate choice by Ben
Franklin, electrons, which typically
carry charge, have the negative sign.
•  This leads to a physically correct but
confusing physical view of current:
Negatively charged
electrons moving
opposite direction.
Current
•  The physically incorrect but perfectly
acceptable view used by most electrical
engineers is instead, this:
Positively charged
“mystery particles”
moving same direction.
Current
•  I recommend you think of positive
charge carriers traveling in the
direction of the current in this class.
Robert R. McLeod, University of Colorado
26
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
The Resistor
Current
Package
+
Potential -
Physics
Drude model
Positive charge
Charge path
Points of collision
Symbol
Note how symbol is similar to charge path.
Robert R. McLeod, University of Colorado
27
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Ohm’s Law
The definition of resistance
•  Friction of the charge carrier travelling
through the solid slows it down.
•  An empirical observation that holds for
many materials under a large range of
conditions is…
I
Ohm’s Law: V = I R
G=
1
R
V
•  Resistance R in ohms [Ω] is
proportionality between V and I.
•  Occasionally useful to use the inverse
quantity, called Conductance, G,
measured in mhos [ ] (Formerly. Now
siemens [S] is recommended, but is not
as much fun).
Ω
Robert R. McLeod, University of Colorado
28
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Power dissipated in a resistor
•  There is a potential drop of 10 V = 10 J/C
across the resistor.
•  5 A, so 5 C of charge is experiencing this loss
of energy per second (A=C/s)
•  The amount of electrical energy lost to heat in
the resistor per unit time is thus
P[sJ = W ] = V [CJ ]I [Cs ] = 50[W ]
•  Resistors have maximum
power ratings:
Robert R. McLeod, University of Colorado
http://denethor.wlu.ca/pc200/resistance/resistors_fixed.shtml 29
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Combining Ohm’s Law
and power equation
= (IR )I
P = VI
= V (V R )
=I R
=V R
P = VI
2
2
The quadratic dependence of dissipated power on voltage
is a major motivation for low-voltage logic families.
Power on the IV diagram:
-
+
Providing
-
+
Dissipating
P [Watts] =
4
2
½
P
I=
V
-
+
Dissipating
Robert R. McLeod, University of Colorado
+
-
Providing
30
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Resistors in series
Equivalent R and voltage division
Q: What is voltage across individual resistors in series?
Step 1: Find equivalent resistance
REq = R1 + R2 = 2 [Ω] + 8 [Ω] = 10 [Ω]
Step 2: Find common current via Ohm’s law
I = VSource REq = 10 [V ] 10 [Ω] = 1 [A ]
Step 3: Find voltage drop on each resistor via Ohm’s law
V1 = IR1 = 1 [A ]2 [Ω] = 2 [V ], V2 = IR2 = 1 [A ]8 [Ω] = 8 [V ]
In general, combine all three to get voltage divider equations
V1 = IR1 = (Vsource REq )R1 = Vsource
Robert R. McLeod, University of Colorado
R1
R1 + R2
31
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Potentiometer
A variable voltage divider
Package
Operation
A
W
B
http://en.wikipedia.org/wiki/Potentiometer
http://www.markallen.com/teaching/ucsd/147a/lectures/lecture3/1.php
Symbol
Equivalent circuit
A
B
W
RAB = RAW + RWB
Robert R. McLeod, University of Colorado
32
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Resistors in parallel
Equivalent R and current division
Q: What is the current through individual resistors in parallel?
Step 1: Find equivalent resistance
or
REq =
(
1
R1
+
) =(
1 −1
R2
1
2
+ 101 ) = 1.6 [Ω]
−1
GEq = G1 + G2
Step 2: Find common current via Ohm’s law
I = VSource REq = 10 [V ] 1.6 [Ω] = 6.25 [A ]
Step 3: Find current through each resistor via Ohm’s law
I1 = Vsource R1 = 10 [V ] 2 [Ω] = 5 [A ], I 2 = Vsource R1 = 10 [V ] 8 [Ω] = 1.25 [A ]
In general, combine all three to get current divider equations
Vsource IREq
1
I1 =
=
=I
R1
R1
R1
Robert R. McLeod, University of Colorado
(
1
R1
+
)
1 −1
R2
=I
G1
G1 + G2
33
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Bonus
Input and output resistance
You design a great
voltage divider to
produce 10V from
your 15 V power
supply.
…but when you hook
up the next
component, the
voltage drops to 6V.
The problem is that
the load resistance is
too low, pulling
current from your
voltage divider. A
high load resistance
works well.
Robert R. McLeod, University of Colorado
34
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 3.1 – Simple Ohm’s Law
Q: The voltage across and current
through R1 are:
A: V = 5 mV, I = 5 mA
B: V = 5 mV, I = 5 A
C: V = 5 V, I = 5 kA
D: V = 5 V, I = 5 mA I = V / R = 5 /1000
E: V = 5 kV, I = 5 A
Robert R. McLeod, University of Colorado
35
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 3.2 – Power
Q: The battery burns out on your
night-light. You replace the battery
with one whose voltage is twice that
of the previous battery. The optical
power emitted by the light bulb…
A: is a quarter what it was before.
B: is half what it was before.
C: is the same as it was before.
D: is twice what it was before.
E: is 4x what it was before. P = V 2
Robert R. McLeod, University of Colorado
R
36
• Lecture 3: Resistance
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 3.3 – Parallel resistors
Q: A resistor R2 of unknown
resistance is added in parallel with
R1. The current drawn from the
battery…
A: goes down.
B: goes down or stays the same.
C: goes up or stays the same.
It could stay the same if R is infinite.
D: goes up.
Yes, it’s a fine point, but a better answer than D.
E: There is not enough info to say.
2
REq =
(
1
R1
+
)
1 −1
R2
Robert R. McLeod, University of Colorado
≤ R1 so I =
V
R Eq
≥
V
R1
37